Pressure inside two soap bubbles are 1.01 and 1.02 atmosphere, respectively. The radio of their volumes is :

According to question, excess pressure inside the soap bubbles,
$\begin{array}{l}{\mathrm{\Delta P}}_1={\mathrm{P}}_1-{\mathrm{P}}_0=0.01=\frac{4\mathrm{T}}{{\mathrm{R}}_1}\to \left(\text{ i }\right)\\ \text{ and }{\mathrm{\Delta P}}_2={\mathrm{P}}_2-{\mathrm{P}}_0=0.02=\frac{4\mathrm{T}}{{\mathrm{R}}_2}\to \left(\text{ ii }\right)\end{array}$
Dividing equation (ii) by (i)
$\frac{1}{2}=\frac{{\mathrm{R}}_2}{{\mathrm{R}}_1}\Rightarrow {\mathrm{R}}_1=2{\mathrm{R}}_2$
Volume: $\mathrm{V}=\frac{4}{3}{\mathrm{\pi R}}^3$

$\Rightarrow \frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}=\frac{{{\mathrm{R}}_1}^3}{{{\mathrm{R}}_2}^3}=\frac{8{{\mathrm{R}}_2}^3}{{{\mathrm{R}}_2}^3}=\frac{8}{1}$